Functional Analysis 8 -- Baire's Category Theorem Application 1

This is course note from Siran Li-Shanghai Jiao Tong University.

Another characterization of meagre sets as a game.

Banach-Mazur Game: Setting: Consider Player Alice and Bob. Given a set \(X \subset [0,1],\) they alternatively take open intervals, s.t. \(A_1 \supset B_1 \supset A_2 ...\)

Then form Y = \(\cap A_i = \cap B_i.\) If \(Y \cap X \neq \phi,\) then Alice wins. Otherwise, Bob wins.

Theorem:

Bob has a winning strategy iff \(X\) is meagre.

Remark: \(\exists B \subset [0,1]\)s.t. \(\forall 0\le a<b\le1,\) \(B \cap [a,b]\) is neither meagre nor residual in \([a,b]\). For this "Brenstein set", neither Alice nor Bob has a winning strategy.

Alice winning strategy is when there exists a set \(I\) such that \(X \cap I\) is residual in \(I.\)

proof: \(\Leftarrow\) If \(X\) is meagre, then \(X \subset \cup M_n\) closed with \(int M_n = \phi.\) So if Bob takes \(B_n \subset [0,1]\backslash M_n,\) then \(Y = \cap B_n\) will miss \(X.\) (all open sets in [0,1] are union of intervals.) Hence Bob wins.

\(\Rightarrow\) Let us enumerate open intervals with rational endpoints by \(I_1, I_2, ...\) Consider "Finite Moves".

Defined \(B_1^{(1)}, B_1^{(2)},...\) inductively: \(B_1^{(1)}\) chosen by Bob according to his winning strategy, if \(A_1 = I_1.\)

Take \(I_2 \cap B_1^{(1)} = \phi.\) Take \(B_1^{(2)}\) as the set \(B_1\) chose by Bob if \(A_1 = I_2.\)

Note that \(B_1^{(2)} \cap B_1^{(1)} = \phi.\) ...

Suppose \(I_k\) is chosen s.t. its disjoint from \(B_1^{(1)}, B_1^{(2)},...B_1^{(k-1)}.\) Then choose \(B_1^{(k)}\) as the set \(B_1\) if \(A_1 = I_k.\) Then \(B_1^{(k)}\) is disjoint from \(B_1^{(1)}, B_1^{(2)},...B_1^{(k-1)}\) by construction.

Then \(\cup B_1^{(k)} = U_1\) is open and dense(otherwise, it will miss an interval in \([0,1]\), which means it misses an \(I_r\). However, we have \(B_1^{(r+1)} \subset \cup B_1^{(k)}\) ).

"Second Move"

By a similar argument with \(B_1^{(i)}\) in place of \([0,1]\) for the first move.

...

By repeating the process we get \(B_2^{(i,j)}\) open intervals s.t. \(B_2^{(i,j)}\) is dense in \(B_1^{(i)}.\)

Thus, \(B_2^{(i,j)}\) is a sequence of disjoint open intervals s.t. \(\cup_{i,j} B_2^{(i,j)}\) is dense in \([0,1].\)

Repeat this process for any \(n \in \mathbb{N}.\) We get \(B_n^{(i)}\) open, disjoint s.t. \(\cup B_n^{(i)} = U_n\) is dense in \([0,1].\)

By Baire's Category theorem, \(Z=\cap U_n\) is residual. Therefore, \(\forall x \in Z,\) it lies in a sequence \(B_1^{i_1} \supset B_2^{i_2} \supset...\) chosen via the winning strategy. So \(x \notin X.\) Thus \(X \subset [0,1]\backslash Z\) is meagre.

GTM John Oxtoby Measure and Category Chapter 6.